10.37 HW 3
Spring 2007
Problem 1. A CSTR of volume 0.602 liters (constant density liquid phase) is operated in
which the following reaction occurs:
A+ B→C+ D
The feed rate of A is 1.16 liters/hr of a solution at concentration 5.87 mmol/L. The feed rate of B is 1.20 liters/hr at a concentration of 38.9 mmol/L. The outlet concentration of species A is 1.094 mmol/L. Calculate the rate constant for this reaction assuming a mass-action rate law of the form: r = k[ A][B] .
First draw a diagram of the problem:
qA [A]0
qout [A] [B]
qB [B]0
The only unknown quantities in the above figure are [B] and
qtot.
A volume balance on the carrier solvent gives the following
relationship (constant fluid density):
q A + q B = qout = 1.16 L/h + 1.20 L/h = 2.36 L/h
Define the extent of reaction:
ξ& = rV [=]
mol
time
A material balance on component A yields the following:
dn A
= FA0 − FA − ξ& = [ A]0 q A − [ A]qout − ξ& .
dt
Setting the accumulation term to zero (steady state
operation) and solving for the reaction rate we find:
Cite as: William Green, Jr., and K. Dane Wittrup, course materials for 10.37 Chemical and Biological Reaction Engineering,
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ξ& = [ A]0 q A − [ A]qout
ξ& = (5.87mmol/L)(1.16L/h ) − (1.094mmol/L )(2.36L/h ) = 4.23mmol/h
A material balance on component B yields the following:
dnB
= FB0 − FB − ξ& = [B]0 q B − [B]qout − ξ& .
dt
Setting accumulation to zero, and solving for [B] we find:
[B] =
[B]0 q B − ξ&
. qout
[B] =
(38.9 mmol/L)(1.20 L/h ) − (4.23 mmol/h )
= 18.0 mmol/L
(2.36 L/h )
Using the given rate law and knowing the extent of
reaction, we can now calculate the rate constant:
ξ& = rV = k[A][B]V
ξ&
k=
k=
[ A][B]V
(4.23 mmol/h)
L
= 0.357
= 357M −1h −1
(1.094 mmol/L)(18.0 mmol/L)(0.602 L )
h mmol
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Problem 2. Consider the catalyzed reaction:
A+ B → B+C
with the second-order rate constant 1.15 x 10-3 m3/mol/ksec. The rate law is
r = k[ A][B] .
What volume of CSTR would be necessary to give 40% conversion of species A if the
feed concentration of A is 96.5 mol/m3, the feed concentration of B is 6.63 mol/m3, and
the flow rate is 0.5 m3/ksec?
First draw a diagram of the problem:
q [A]0 [B]0
[A] [B] [C]
Define the extent of reaction:
mol
ξ& = rV[=]
time
Define the conversion in terms of the variables of the
problem:
XA =
mol A reacted
mol A fed
=
− rAV
q[ A]0
=
ξ&
q[ A]0
A material balance on species A gives the following
equation:
dn A
= FA0 − FA − ξ& = [A]0 q − [ A]q − ξ&
dt
At steady state:
0 = [ A]0 q − [ A]q − ξ&
ξ& = [ A]0 q − [A]q
Thus our conversion definition is equivalent to:
Cite as: William Green, Jr., and K. Dane Wittrup, course materials for 10.37 Chemical and Biological Reaction Engineering,
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XA =
[ A]0 q − [ A]q
[ A]
=1−
q[ A]0
[ A]0
[ A] = (1 − X A )[ A]0
A material balance on species [B] gives the following:
dnB
= FB0 − FB + 0ξ& = [B]0 q − [B]q
dt
At steady state:
[B] = [B]0 .
Plugging in the rate expression in the extent of reaction,
ξ& = rV = k[ A][B]V ,
the steady state A balance becomes:
0 = [ A]0 q − [ A]q − ξ& = ([ A]0 − [ A])q − k[ A][B]V .
Using the expressions derived for [B] and [A] in terms of
given values, we find:
0 = [ A]0 X A q − k[ A]0 (1 − X A )[B]0 V
Solving the equation for V we find:
V =
V
[ A]0 X A q
X Aq
=
k[ A]0 (1 − X A )[B]0 k (1 − X A )[B]0
(0.40)(0.5 m 3 /ksec)
=
= 43.7 m 3
3
3
−3
(1.15 ×10 m /(mol ⋅ ksec))(1 − 0.40)(6.63 mol/m )
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Problem 3. Two configurations of CSTRs are contemplated for performing reversible
hydrolysis of compound A to produce compounds B and C. The forward reaction is
pseudo-first-order with respect to A, with rate constant k1= 1.82 x 10-4 s-1. The reverse
reaction is second-order with rate constant k-1 = 4.49 x 10-4 M-1s-1 .
(1) A → B + C; r1 = k1[ A]
(2) B + C → A; r2 = k −1[B][C]
The feed is a dilute aqueous solution of A (concentration 0.25 mol/L) at a rate of 0.25
liters per hour.
Consider the following two configurations:
a) a single 15 liter CSTR.
b) three 5-liter CSTRs in series, with 75% of product species B & C selectively removed
between stages 1 and 2 and between stages 2 and 3, with appropriate adjustment in flow
rate; the volumetric flow rates in the two streams leaving a separator are proportional to
the total number of moles of A, B, and C in each stream. See the separator diagram
below.
q0 FA0 FB0 FC0
q1 FA1 FB1 FC1
Separator
q2 FA2 FB2 FC2
If q is the volumetric flow rate and F is the molar flow rate, then the separator follows the
relationships:
FA 2 = 0
FB 2 = 0.75FB 0
FC 2 = 0.75FC 0
F + FB1 + FC1
q1
= A1
q2 FA2 + FB2 + FC 2
A full flow diagram is shown below.
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Determine the steady-state production of compound B in mol/h for options a) and b).
First, consider the 1-reactor case. Draw a diagram of the
situation (the control volume).
q0 [A]0
[B]0=[C]0=0
V=15 L
[A] [B] [C]
Define extents of reaction:
ξ&1 = r1V = k1[ A]V
ξ&2 = r2V = k −1[ B][C ]V
Steady state A, B, and C material balances give the
following equations:
0 = FA0 − FA − ξ&1 + ξ&2
0 = F − F + ξ& − ξ&
B0
B
1
2
0 = FC 0 − FC + ξ&1 − ξ&2
These three equations can be written in terms of only three
variables: [A], [B], and [C]. The system is fully
specified. We must solve these equations simultaneously.
The algebra is easier if we combine equations creatively to
make them simpler. When you combine two equations, you
keep the new equation and only one of the old ones, just as
is done in linear algebra.
Notice that:
FB0 = FC 0 = 0 .
Cite as: William Green, Jr., and K. Dane Wittrup, course materials for 10.37 Chemical and Biological Reaction Engineering,
Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
Hence, subtracting the B balance equation from the C
balance equation gives the result:
FB = FC
q0 [ B] = q0 [C ]
[ B] = [C ]
Let this new equation replace the C mole balance.
The equality of concentrations of C and B is true whenever
[C]0 and [B]0 are equal. [B]0 has been intentionally left
as a variable so that the same equations derived here will
also apply to the reactors in part b, where [B]0 will not
necessarily equal zero.
Adding together the B and A balance gives the following
equation:
0 = FA0 − FA + FB0 − FB = q([ A]0 − [ A] + [ B]0 − [ B])
Dividing by q0 and solving for [A] we get:
[ A] = ([ A]0 + [ B]0 − [ B])
Let this new equation replace the B mole balance.
Now we can write the A balance as a single equation with a
single variable, [B] using our two new equations:
0 = FA0 − FA − ξ&1 + ξ&2
0 = q0 ([ A]0 − [ A]) − k1V [ A] + k −1V [ B][C ]
0 = q0 ([ A]0 − ([ A]0 + [ B]0 − [ B ])) − k1V ([ A]0 + [ B]0 − [ B]) + k −1V [ B][ B]
0 = q0 [ B] − q0 [ B]0 − k1V [ A]0 − k1V [ B]0 + k1V [ B] + k −1V [ B]2
0 = (k −1V )[ B]2 + (q0 + k1V )[ B] + (− q0 [ B]0 − k1V [ A]0 − k1V [ B ]0 )
0 = a[ B ]2 + b[ B ] + c
a = (k −1V )
Cite as: William Green, Jr., and K. Dane Wittrup, course materials for 10.37 Chemical and Biological Reaction Engineering,
Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
b = (q0 + k1V )
c = (− q0 [ B]0 − k1V [ A]0 − k1V [ B]0 )
This quadratic equation in [B] can be solved with the
quadratic formula:
[ B] =
− b ± b 2 − 4ac
2a
Since the coefficient b will always be POSITIVE, we know
that we have to take the (+) root. The (-) root will
always be negative. Also notice that a is always positive
and c is always negative, so that the descriminant,
(b
2
− 4ac ) > 0 ,
is always positive, and both roots are real.
write:
[ B] =
Hence, we may
− b + b 2 − 4ac
2a
with no fear of getting a negative or imaginary result.
Converting all time to hours, we find that for part a
L2
⎛ 3600s ⎞
a = (k −1V ) = (4.49 ×10 M s )⎜
⎟(15 L ) = 24.246
mol ⋅ h
⎝ h ⎠
−4
-1 −1
L
⎛ 3600s ⎞
b = (q0 + k1V ) = (0.25 L/h ) + (1.82 ×10 −4 s −1 )⎜
⎟(15 L ) = 10.078
h
⎝ h ⎠
c = (− q0 [B]0 − k1V[ A]0 − k1V[B]0 )
mol
⎛ 3600s ⎞
c = 0 − (1.82 ×10 −4 s −1 )⎜
⎟(15 L )(0.25M ) − 0 = −2.457
h
⎝ h ⎠
Plugging the values of the coefficients into the quadratic
equation, we find:
[B] = 0.172 M
Cite as: William Green, Jr., and K. Dane Wittrup, course materials for 10.37 Chemical and Biological Reaction Engineering,
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Hence, the desired production rate of B is:
L⎞
⎛
FB = q0 [B] = ⎜ 0.25 ⎟(0.172 M ) = 0.0430 mol/h
h⎠
⎝
Cite as: William Green, Jr., and K. Dane Wittrup, course materials for 10.37 Chemical and Biological Reaction Engineering,
Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
In order to do part b, it is helpful to notice that
everything flows from left to right, with no recycle loops.
Hence we can solve each reactor and separator in sequence,
and add up the total B recovery at the end of the process.
Also, we could get it from an overall balance (a box around
the entire system) once we knew the final outlet
concentration of A.
For simplicity of notation, let the zero subscript refers
to the current reactor feed; no subscript refers to the
current reactor output.
Reactor 1 q0=0.25L/h [A]0=0.25M [B]0=[C]0=0M V=5 L
[A] [B] [C]
Reactor 1 follows the same relationships as the single
reactor system, but with different volume. Hence the same
material balances apply, and the final equation in terms of
[B] is once again:
0 = (k −1V )[B]2 + (q0 + k1V )[B] + (− q0 [B]0 − k1V[ A]0 − k1V[B]0 )
0 = a[B]2 + b[B] + c
a = (k −1V )
b = (q0 + k1V )
c = (− q0 [B]0 − k1V[ A]0 − k1V[B]0 )
[ B] =
− b + b 2 − 4ac
2a
Plugging in the numbers, we find the quadratic coefficients
to be:
L2
⎛ 3600s ⎞
a = (k −1V ) = 4.49 ×10 −4 M -1s−1 ⎜
⎟(5 L ) = 8.082
mol ⋅ h
⎝ h ⎠
(
)
L
⎛ 3600s ⎞
b = (q0 + k1V ) = (0.25 L/h ) + (1.82 ×10 −4 s−1 )⎜
⎟(5 L ) = 3.526
h
⎝ h ⎠
Cite as: William Green, Jr., and K. Dane Wittrup, course materials for 10.37 Chemical and Biological Reaction Engineering,
Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
c = (− q0 [B]0 − k1V[ A]0 − k1V[B]0 )
mol
⎛ 3600s ⎞
c = 0 − (1.82 ×10 −4 s−1 )⎜
⎟(5 L )(0.25M ) − 0 = −0.819
h
⎝ h ⎠
Plugging the quadratic coefficients into the formula for
[B] we find:
[B] = 0.168 M
Using our formula for [A] in terms of [B], we find:
[ A] = ([ A]0 + [B]0 − [B])
[ A] = 0.25 M + 0 M − 0.168 M = 0.082 M
Cite as: William Green, Jr., and K. Dane Wittrup, course materials for 10.37 Chemical and Biological Reaction Engineering,
Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
Next we proceed to separator 1, carry forward our inputs,
and rename our subscripted variables, so that they refer to
the current separator.
q0=0.25 L/h
[A]0 = 0.082 M
[B]0=[C]0=0.168 M
q1
[A]1
[B]1
Separator
1
q2
[A]2
[B]2
First we need to find the flow rate partitioning, then we
can calculate the new concentrations from an A and B
balance. (Since the separator treats B and C the same,
these concentrations will continue to be equal in all
streams exiting the separator.) The separator equations
are:
FA 2 = 0
FB 2 = 0.75FB 0
FC 2 = 0.75FC 0
F + FB1 + FC1
q1
= A1
q2 FA 2 + FB 2 + FC 2
Hence using the definition of molar flow rate, the
knowledge that [B]0 = [C]0, and the above separator
relationships, we find that:
q1
q2
q1
q2
=
=
(q0 [ A]0 ) + 0.25(q0 [B]0 ) + 0.25(q0 [C]0 ) [ A]0 + 0.5[B]0
=
0 + 0.75(q0 [B]0 ) + 0.75(q0 [C]0 )
1.5[B]0
[ A]0 + 0.5[B]0
1.5[B]0
=
(0.082) + 0.5(0.168)
= 0.659
1.5(0.168)
q1 = 0.659q2
Assuming a constant density liquid phase, we have a volume
balance:
q0 = q1 + q2 = 1.659q2
q2 =
0.25 L/h
1.659
= 0.151 L/h
Cite as: William Green, Jr., and K. Dane Wittrup, course materials for 10.37 Chemical and Biological Reaction Engineering,
Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
q1 = 0.659 × 0.151 L/h = 0.100 L/h
Thus we have recovery of product B at this stage of:
FB1E = 0.75(q0 [B]0 ) = 0.75(0.25 L/h )(0.168 M ) = 0.0315 mol/h
The last things to calculate at this step are the
concentrations that are fed to the next reactor. Using the
mole balance on A:
FA0 = q0 [ A]0 = FA1 = q1[ A]1
[ A]1 =
q0
q1
[ A]0 =
0.25
0.100
0.082
mol
L
= 0.205 mol/L . Using the separator relationships for B and C:
FB1 = 0.25FB 0
[B]1 = 0.25
q0
q1
[B]0 = 0.25
0.25
0.100
0.168
mol
L
= 0.105 mol/L
[ B ]1 = [C ]1
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Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
Next we proceed to reactor 2 using the results from
separator 1 and rename our variables so they refer to the
current reactor.
Reactor 2
q0=0.100L/h
[A]0=0.205 M
[B]0=[C]0=0.105 M
V=5 L
[A] [B] [C]
All the same equations apply with different initial
concentrations and flow rates.
L2
⎛ 3600s ⎞
(
)
a = (k −1V ) = 4.49 ×10 −4 M -1s−1 ⎜
5
L
=
8.082
⎟
mol ⋅ h
⎝ h ⎠
(
)
L
⎛ 3600s ⎞
b = (q0 + k1V ) = (0.100 L/h ) + (1.82 ×10 −4 s−1 )⎜
⎟(5 L ) = 3.376
h
⎝ h ⎠
c = (− q0 [B]0 − k1V[ A]0 − k1V[B]0 )
(
)
L⎞
⎛
⎛ 3600s ⎞
c = ⎜ − 0.100 ⎟(0.105M ) − 1.82 ×10 −4 s −1 ⎜
⎟(5 L )[0.205 + 0.105]M
h⎠
⎝
⎝ h ⎠
mol
c = −1.026
h
The quadratic solution gives:
[B] = 0.204 M
Solving for [A] from the [A]/[B] relationship we find:
[ A] = ([ A]0 + [B]0 − [B])
[ A] = 0.205 M + 0.105 M − 0.204 M = 0.106 M
These concentrations will be fed to separator 2.
Cite as: William Green, Jr., and K. Dane Wittrup, course materials for 10.37 Chemical and Biological Reaction Engineering,
Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
Next we proceed to separator 2 and rename our subscripted
variables.
q0=0.100 L/h
[A]0 = 0.106 M
[B]0=[C]0=0.204 M
q1
[A]1
[B]1
Separator
2
q2
[A]2
[B]2
The flow rate partitioning is given by:
q1
q2
=
[ A]0 + 0.5[B]0
1.5[B]0
=
(0.106) + 0.5(0.204)
= 0.680
1.5(0.204)
Hence, from the constant density volume balance:
q2 =
q0
1.680
=
0.100 L/h
1.680
= 0.0595 L/h
q1 = q0 − q1 = (0.100 − 0.0595)
L
h
= 0.0405 L/h
The amount of [B] recovered is:
FB 2 E = 0.75(q0 [B]0 ) = 0.75(0.100L/h )(0.204 M ) = 0.0153 mol/h
The concentrations of [A], [B], and [C] into the next
reactor are:
FA0 = q0 [ A]0 = FA1 = q1[ A]1
0.100
q
mol
[ A]1 = 0 [ A]0 =
0.106
= 0.262 mol/L . 0.0405
q1
L
FB1 = 0.25FB 0
0.100
q
mol
[B]1 = 0.25 0 [B]0 = 0.25
0.204
= 0.126 mol/L
0.0405
q1
L
[ B ]1 = [C ]1
Lastly, we rename our concentrations and proceed to reactor
3.
Cite as: William Green, Jr., and K. Dane Wittrup, course materials for 10.37 Chemical and Biological Reaction Engineering,
Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
Lastly, we proceed to reactor 3 using the results from
separator 2.
Reactor 3
q0=0.0405L/h
[A]0=0.262 M
[B]0=[C]0=0.126 M
V=5 L
[A] [B] [C]
All the same equations apply with different initial
concentrations and flow rates.
L2
⎛ 3600s ⎞
a = (k −1V ) = 4.49 ×10 −4 M -1s −1 ⎜
⎟(5 L ) = 8.082
mol ⋅ h
⎝ h ⎠
L
⎛ 3600s ⎞
b = (q0 + k1V ) = (0.0405 L/h ) + 1.82 ×10 −4 s −1 ⎜
⎟(5 L ) = 3.3165
h
⎝ h ⎠
c = (− q0 [B]0 − k1V[ A]0 − k1V[B]0 )
(
)
(
(
)
)
L⎞
⎛
⎛ 3600s ⎞
c = ⎜ − 0.100 ⎟(0.126M ) − 1.82 ×10 −4 s −1 ⎜
⎟(5 L )[0.262 + 0.126]M
h⎠
⎝
⎝ h ⎠
mol
c = −1.284
h
The quadratic solution gives:
[B] = 0.243 M
Solving for [A] from the A/B relationship equation we find:
[ A] = ([ A]0 + [B]0 − [B])
[ A] = 0.262 M + 0.126 M − 0.243 M = 0.145 M
Since there is no separator, the amount of B recovered on
this step is just the amount leaving the reactor:
L⎞
⎛
FB3E = [B]q0 = (0.243M)⎜ 0.0405 ⎟ = 0.0098mol/h
h⎠
⎝
Thus the total amount of B recovered by this path is:
FB = FB1E + FB 2 E + FB 3 E = (0.0315 + 0.0153 + 0.0098)
mol
h
= 0.0566
mol
h
Another way to calculate it, and check our consistency, is
to calculate the amount of B recovered from the overall A
and B balances around the whole reaction system (any A that
disappears must be present as B):
Cite as: William Green, Jr., and K. Dane Wittrup, course materials for 10.37 Chemical and Biological Reaction Engineering,
Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
0 = FAin − FAout + FBin − FBout
L⎞
L⎞
mol
⎛
⎛
FBout = FAin − FAout = ⎜ 0.25 ⎟(0.25M ) − ⎜ 0.0405 ⎟(0.145M ) = 0.0566
h⎠
h⎠
h
⎝
⎝
From the agreement of the numbers, it appears that mass was
conserved overall.
Notice that incorporating the selective separation process
to remove product along the way, the amount of B recovered
was improved with the same total volume of reactor.
Due to the truncation error and numerical rounding, accept
any answers within 2% of these values.
FB ,3rctr = (0.0566 ± 0.0011)
FB ,1rctr = (0.0430 ± 0.0009 )
mol
h
mol
h
The modular setup of this problem (reactor units, separator
units) makes a Matlab implementation straightforward. See
included Matlab m-file hw3prob3.
Cite as: William Green, Jr., and K. Dane Wittrup, course materials for 10.37 Chemical and Biological Reaction Engineering,
Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].